Parabola Equation: Focus (3,1), Directrix Y=3

Hey math whizzes and number crunchers! Today, we're diving deep into the fascinating world of parabolas. You know, those U-shaped curves that pop up everywhere from satellite dishes to the trajectory of a thrown ball. We're going to tackle a specific problem: finding the equation of a parabola with a focus at $(3,1)$ and a directrix at $y=3$. This isn't just about memorizing formulas, guys; it's about understanding the fundamental definition of a parabola and how that definition translates into an algebraic equation. By the end of this, you'll be able to solve similar problems with confidence, impressing your friends and acing your next math test. So, grab your notebooks, sharpen those pencils, and let's get this parabola party started!

Understanding the Building Blocks: Focus and Directrix

Before we jump into solving, let's get crystal clear on what the focus and directrix are. Think of them as the two key ingredients that define a parabola. The focus is a fixed point inside the parabola. The directrix is a fixed line outside the parabola. The magic of a parabola lies in its definition: a parabola is the set of all points that are equidistant from the focus and the directrix. This means if you pick any point on the parabola, the distance from that point to the focus is exactly the same as the distance from that point to the directrix. This core concept is the foundation upon which we'll build our equation. It's like the DNA of the parabola, dictating its shape and position. So, when you see a focus and a directrix, immediately think about this equal-distance property. It's your golden ticket to unlocking the parabola's equation.

Visualizing Our Parabola

Let's visualize the given information: our focus is at the point $(3,1)$, and our directrix is the horizontal line $y=3$. First, let's find the vertex. The vertex of a parabola is always located exactly halfway between the focus and the directrix. Since the focus is at $y=1$ and the directrix is at $y=3$, the midpoint in the y-direction is $(1+3)/2 = 2$. The x-coordinate of the focus is 3, and since the directrix is a horizontal line, the x-coordinate of the vertex will also be 3. Therefore, the vertex of our parabola is at $(3,2)$.

Now, consider the orientation. The focus $(3,1)$ is below the directrix $y=3$. This tells us that our parabola must open downwards. Remember, parabolas open towards their focus. If the focus were above the directrix, it would open upwards. If the focus were to the left or right of a vertical directrix, it would open left or right, respectively. The distance from the vertex $(3,2)$ to the focus $(3,1)$ is $2-1=1$. This distance is often denoted by 'p', so $p=1$. Since the parabola opens downwards, we'll be using a negative sign in our final equation, reflecting this downward orientation.

Deriving the Equation: The Distance Formula in Action

Alright guys, here's where the math really comes alive. We're going to use the definition of the parabola – that the distance from any point $(x,y)$ on the parabola to the focus $(3,1)$ is equal to the distance from $(x,y)$ to the directrix $y=3$. Let's break it down:

1. Distance from $(x,y)$ to the focus $(3,1)$: We use the distance formula: $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. So, the distance is $\sqrt{(x-3)^2 + (y-1)^2}$.

2. Distance from $(x,y)$ to the directrix $y=3$: Since the directrix is a horizontal line $y=3$, the distance from any point $(x,y)$ to this line is simply the absolute difference in the y-coordinates: $|y-3|$.

3. Equating the distances: According to the definition of a parabola, these two distances must be equal: $\sqrt{(x-3)^2 + (y-1)^2} = |y-3|$

4. Squaring both sides to eliminate the square root: To simplify, we square both sides of the equation: $(x-3)^2 + (y-1)^2 = (y-3)^2$

5. Expanding and simplifying: Now, let's expand the squared terms involving $y$: $(x-3)^2 + (y^2 - 2y + 1) = (y^2 - 6y + 9)$

Notice that $y^2$ appears on both sides, so we can cancel it out: $(x-3)^2 - 2y + 1 = -6y + 9$

Our goal is to isolate $y$ to get the equation in the form $y = ax^2 + bx + c$ or similar. Let's move the terms involving $y$ to one side and the constants to the other: $(x-3)^2 + 1 - 9 = -6y + 2y$ $(x-3)^2 - 8 = -4y$

6. Solving for $y$: Now, divide both sides by -4 to get $y$ by itself: $y = \frac{(x-3)^2 - 8}{-4}$ $y = -\frac{1}{4}(x-3)^2 + \frac{8}{4}$ $y = -\frac{1}{4}(x-3)^2 + 2$

And there you have it! The equation of the parabola that perfectly fits our focus and directrix is $y = -\frac{1}{4}(x-3)^2 + 2$. This matches option C. Pretty neat, huh? We used the fundamental definition and a bit of algebraic elbow grease to arrive at the answer. Remember this process, and you'll be a parabola pro in no time!

Connecting to the Standard Form: Vertex Form

The equation we derived, $y = -\frac{1}{4}(x-3)^2 + 2$, is in what we call vertex form. The general vertex form of a parabola that opens upwards or downwards is $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex of the parabola. In our case, we found the vertex to be $(3,2)$. Plugging this into the vertex form, we get $y = a(x-3)^2 + 2$.

The coefficient '$a$' determines how wide or narrow the parabola is and also its direction of opening. If $a > 0$, the parabola opens upwards. If $a < 0$, it opens downwards. We already established that our parabola opens downwards because the focus $(3,1)$ is below the directrix $y=3$. So, we know $a$ must be negative.

Now, let's relate this to the distance 'p' we talked about earlier. The relationship between '$a$' and '$p$' for parabolas opening vertically is $a = \frac{1}{4p}$. We found that the distance from the vertex $(3,2)$ to the focus $(3,1)$ is $p=1$. Plugging this into the formula for $a$: $a = \frac{1}{4(1)} = \frac{1}{4}$.

However, since our parabola opens downwards, we need to incorporate that negative sign. So, $a = -\frac{1}{4}$.

Substituting this value of $a$ back into the vertex form $y = a(x-3)^2 + 2$, we get: $y = -\frac{1}{4}(x-3)^2 + 2$

This confirms our result from the distance formula derivation. It's always great when two different methods yield the same answer! This vertex form is super useful because it immediately tells you the vertex's coordinates and the parabola's direction of opening. It's like a cheat sheet for the parabola's key features.

Why the Other Options Don't Work

Let's take a quick look at why the other multiple-choice options are incorrect. This helps solidify our understanding.

• Option A) $y=-1/4(x-3)^2+3$: This equation has the correct 'a' value (-1/4) indicating a downward opening parabola, and the $(x-3)^2$ term suggests the x-coordinate of the vertex is 3. However, the $+3$ term indicates the vertex's y-coordinate is 3. We calculated the vertex to be $(3,2)$, not $(3,3)$. If the vertex were $(3,3)$, the focus would be at $(3, 3 - 1/4) = (3, 11/4)$ and the directrix would be $y = 3 + 1/4 = 13/4$, which doesn't match our given focus and directrix.

• Option B) $y=1/4(x-3)^2+2$: This equation has the correct vertex y-coordinate (2) and the x-coordinate of the vertex is 3, so the vertex is $(3,2)$. However, the coefficient is $+1/4$, meaning this parabola opens upwards. Our focus $(3,1)$ is below the directrix $y=3$, so the parabola must open downwards. An upward-opening parabola with vertex $(3,2)$ would have its focus above the vertex.

• Option D) $y=1/4(x-3)^2+3$: This option suffers from two errors. First, the coefficient $1/4$ implies an upward opening parabola, which contradicts our given focus and directrix. Second, the $+3$ term implies a vertex y-coordinate of 3, which is incorrect; the vertex is at $y=2$.

By analyzing the vertex form and the sign of the coefficient '$a$', we can quickly eliminate incorrect options and confirm our correctly derived equation. It's all about understanding what each part of the equation tells you about the parabola's properties!

Conclusion: Mastering Parabolas

So there you have it, folks! We've successfully found the equation of a parabola given its focus at $(3,1)$ and its directrix at $y=3$. We achieved this by leveraging the fundamental definition of a parabola – the locus of points equidistant from the focus and directrix – and applying the distance formula. We then simplified the resulting equation to arrive at $y = -\frac{1}{4}(x-3)^2 + 2$. We also explored the vertex form of the parabola, $y = a(x-h)^2 + k$, and saw how our derived equation fits perfectly, with the vertex at $(3,2)$ and the coefficient $a = -\frac{1}{4}$ correctly indicating a downward opening shape.

Understanding the relationship between the focus, directrix, vertex, and the coefficient '$a$' is key to mastering parabola problems. Remember that the vertex is always halfway between the focus and the directrix, and the sign of '$a$' tells you whether the parabola opens upwards ($a>0$) or downwards ($a<0$). The magnitude of '$a$' dictates the 'width' of the parabola – a larger absolute value of '$a$' makes it narrower, while a smaller absolute value makes it wider.

This problem is a fantastic example of how abstract mathematical concepts can be translated into concrete equations. Whether you're sketching graphs, solving physics problems, or designing optical equipment, understanding parabolas is a valuable skill. Keep practicing, keep exploring, and don't be afraid to tackle more challenging problems. You've got this! If you enjoyed this deep dive, keep checking back to Plastik Magazine for more math adventures.